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KIM [24]
3 years ago
14

Angle α lies in quadrant II , and tanα=−125 . Angle β lies in quadrant IV , and cosβ=35 .

Mathematics
1 answer:
Artist 52 [7]3 years ago
7 0

Answer:

cos(\alpha+\beta)=\frac{33}{65}

Step-by-step explanation:

step 1

Find cos α

we know that

tan^2(\alpha)+1=sec^2(\alpha)

we have

tan(\alpha)=-\frac{12}{5}

substitute

(-\frac{12}{5})^2+1=sec^2(\alpha)

sec^2(\alpha)=\frac{144}{25}+1

sec^2(\alpha)=\frac{169}{25}

sec(\alpha)=\pm\frac{13}{5}

Remember that Angle α lies in quadrant II

so

sec α is negative

sec(\alpha)=-\frac{13}{5}

Find the value of cos α

cos)\alpha)=\frac{1}{sec(\alpha)}

so

cos(\alpha)=-\frac{5}{13}

step 2

Find sin α

we know that

tan(\alpha)=\frac{sin(\alpha)}{cos(\alpha)}

sin(\alpha)=tan(\alpha)cos(\alpha)

we have

tan(\alpha)=-\frac{12}{5}

cos(\alpha)=-\frac{5}{13}

substitute

sin(\alpha)=(-\frac{12}{5})(-\frac{5}{13})

sin(\alpha)=\frac{12}{13}

step 3

Find sin β

we know that

sin^2(\beta)+cos^2(\beta)=1

we have

cos(\beta)=\frac{3}{5}

substitute

sin^2(\beta)+(\frac{3}{5})^2=1

sin^2(\beta)=1-(\frac{3}{5})^2

sin^2(\beta)=1-\frac{9}{25}

sin^2(\beta)=\frac{16}{25}

sin(\beta)=\pm\frac{4}{5}

Remember that

Angle β lies in quadrant IV

so

sin β is negative

sin(\beta)=-\frac{4}{5}

step 4

Find cos(α−β)

we know that

cos(\alpha+\beta)=cos(\alpha)cos(\beta)-sin(\alpha)sin(\beta)

we have

cos(\alpha)=-\frac{5}{13}

cos(\beta)=\frac{3}{5}

sin(\alpha)=\frac{12}{13}

sin(\beta)=-\frac{4}{5}

substitute the given values

cos(\alpha+\beta)=(-\frac{5}{13})(\frac{3}{5})-(\frac{12}{13})(-\frac{4}{5})

cos(\alpha+\beta)=(-\frac{15}{65})+(\frac{48}{65})

cos(\alpha+\beta)=\frac{33}{65}

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